(b)Let specific gravities of concrete and saw dust are ${\rho _1}$ and ${\rho _2}$ respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere
$\frac{4}{3}\pi ({R^3} – {r^3}){\rho _1}g + \frac{4}{3}\pi {r^3}{\rho _2}g = \frac{4}{3}\pi {R^3} \times 1 \times g$
==> ${R^3}{\rho _1} – {r^3}{\rho _1} + {r^3}{\rho _2} = {R^3}$
==> ${R^3}({\rho _1} – 1) = {r^3}({\rho _1} – {\rho _2})$ ==> $\frac{{{R^3}}}{{{r^3}}} = \frac{{{\rho _1} – {\rho _2}}}{{{\rho _1} – 1}}$
==> $\frac{{{R^3} – {r^3}}}{{{r^3}}} = \frac{{{\rho _1} – {\rho _2} – {\rho _1} + 1}}{{{\rho _1} – 1}}$
==> $\frac{{({R^3} – {r^3}){\rho _1}}}{{{r^3}{\rho _2}}} = \left( {\frac{{1 – {\rho _2}}}{{{\rho _1} – 1}}} \right)\;\frac{{{\rho _1}}}{{{\rho _2}}}$
==> $\frac{{{\rm{Mass of concrete }}}}{{{\rm{Mass of saw dust}}}} = \left( {\frac{{1 – 0.3}}{{2.4 – 1}}} \right) \times \frac{{2.4}}{{0.3}} = 4$